Theorem fneq1i 5478

Description: Equality inference for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq1i.1	|- F = G

Assertion

Ref	Expression
fneq1i	|- ( F Fn A <-> G Fn A )

Proof of Theorem fneq1i

Step	Hyp	Ref	Expression
1		fneq1i.1	. 2 |- F = G
2		fneq1 5472	. 2 |- ( F = G -> ( F Fn A <-> G Fn A ) )
3	1, 2	ax-mp 5	1 |- ( F Fn A <-> G Fn A )

This proof depends on syntax axioms:   <-> wb 105   = wceq 1402   Fn wfn 5375

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-sn 3715  df-pr 3716  df-op 3718  df-br 4132  df-opab 4194  df-rel 4783  df-cnv 4784  df-co 4785  df-dm 4786  df-fun 5382  df-fn 5383

This theorem is used by:  fnunsn  5493  fnopabg  5510  f1oun  5662  f1oi  5682  f1osn  5684  ovid  6208  tfri1d  6610  frec2uzrand  10863  frec2uzf1od  10864  frecfzennn  10884  xnn0nnen  10895  prdsinvlem  14303  dfrelog  16089  edgstruct  16510  nninfsellemeqinf  17265